MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a smooth projective variety over $\mathbb{C}$, and let $\alpha \in H^{2k}(X)$ be an algebraic cohomology class. Let us fix an $l$ such that $H^l(X) \neq 0$ and $H^{l+2k}(X) \neq 0$.

The general question (perhaps a bit vague) is: does anybody know a sufficient condition on $X$ such that the intersection map

$ \alpha \cup - : H^l(X) \to H^{l+2k}(X) $

is NOT the zero map?

In particular, I would be interested in one of the following two cases.

1) Both $h^l$ and $h^{l+2k}$ can be arbitrarily large (so that being the zero map is very unlikely, so to speak).

2) The number $2l+2k$ equals $2 \dim_{\mathbb{C}}(X)$, thus $H^l\cong H^{l+2k}$ by Poincarè duality.

Would it help to assume that $\alpha$ is the Euler class (top Chern class) of a vector bundle on $X$?

ADDED: I am interested in the case when $l$ is odd.

share|cite|improve this question
I merged your "algebraic" and "geometry" tags. – Donu Arapura May 3 '12 at 20:57
Thanks! So I could add algebraic-topology. – calc May 4 '12 at 17:47

Since you ask only for a sufficient condition, let $l=2n$ be even and suppose that $\alpha$ is effective i.e. a positive linear combination of fundamental classes of subvarieties $V_i$. Let $[H]\in H^2(X)$ be the fundamental class of a hyperplane section. Then $\beta=\[H]^n\in H^l(X)$ is a class such that $\alpha\cup \beta\not=0$. To see this, note that after cupping with additional copies of $[H]$, we obtain the (positive weighted) sum of degrees of $V_i$.

Addendum (added in response to the edited question): The case where $l$ is odd is actually more interesting. Let me give an example to show that the desired result can fail without additional hypotheses. Choose a smooth projective variety $X'$ with $\dim X'=n>1$ and $H^l(X')\not=0$ with $l$ odd. Now blow up a smooth codimension $k+1>1$ subvariety $V$ to get $X$. Then $H^l(X')\subset H^l(X)$, so it's also nonzero. Let $E$ be the exceptional divisor. This is a $\mathbb{P}^{k}$ bundle over $V$. Let $\alpha=[\mathbb{P}^k]$ (one of the fibres). This is a nonzero class, but $\alpha\cup:H^l(X)\to H^{l+2k}(X)$ is zero, because it factors through restriction to $\mathbb{P}^{k}$. Note that $\dim H^l$ and $\dim H^{l+2k}$ can be arbitrarily large.

share|cite|improve this answer
Thank you very much for your answer. Actually the problem where my question arises has $l$ odd. I forgot to specify this in the formulation, I apologize for this. – calc May 4 '12 at 17:50
Very nice example, thanks. – calc May 6 '12 at 8:07

Also, have a look at the Hard Lefschetz Theorem, if you have not seen it yet.

share|cite|improve this answer
Thank you very much for your answer. The point here is that I do not have much control on $\alpha$, and so I do not knot how to relate it to the hyperplane (or kahler) class. Is there some trick (or philosophy) to circumvent this problem? – calc May 4 '12 at 17:53
Do you know if it is a top Chern class of some vector bundle? – Vladimir Baranovsky May 7 '12 at 4:09
Yes, $\alpha$ is the top Chern class of a vector bundle. Does this help? – calc May 9 '12 at 6:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.